TITLE: Counterexample of Compactness
QUESTION [5 upvotes]: Let $X$ be a metric space and $E\subset X$.
Let {$G_i$} be an open cover of $E$

For every open cover {$G_i$}, there exists a finite subcover {$G_{i_n}$} of $E$ such that $G_{i_n} \in${$G_i$}.
For every open cover {$G_i$}, there exists {$M_n$}, a finite family of open sets, such that $E\subset$$\bigcup M_n \subset \bigcup G_i$.

As you know, if 1 is true, then $E$ is compact.
I think 1 and 2 both have the same meaning, but can't prove the equivalence. (1→2 is trivial, but 2→1 is not)
If 1&2 are not equivalent, please give me some counterexamples..

REPLY [5 votes]: $2$ does not imply $1$. Take for example $E=M=\mathbb{R}$, with the open cover $\{ G_n=(-\infty,n)|n \in \mathbb{N}\}$. This cover has no finite subcover, but we can just take $M=\mathbb{R}$ and $\{ M \}$ is a set satisfying the conclusion of $2$.